Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
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- Product of series
-
How to use it?
- Sum of series:
-
(3-2i-5i^2+8i^3)
- ln(1-1/k^2)
-
e^n/(1+e^(2*n))
-
cos(npi)/5^n
-
Identical expressions
- cos(npi)/ five ^n
- co sinus of e of (n Pi ) divide by 5 to the power of n
- co sinus of e of (n Pi ) divide by five to the power of n
- cos(npi)/5n
- cosnpi/5n
- cosnpi/5^n
- cos(npi) divide by 5^n
Sum of series cos(npi)/5^n
The solution
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[src]
oo ____ \ ` \ cos(n*pi) \ --------- / n / 5 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\cos{\left(\pi n \right)}}{5^{n}}$$
Sum(cos(n*pi)/5^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\cos{\left(\pi n \right)}}{5^{n}}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \cos{\left(\pi n \right)}$$
and
$$x_{0} = -5$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-5 + \lim_{n \to \infty} \left|{\frac{\cos{\left(\pi n \right)}}{\cos{\left(\pi \left(n + 1\right) \right)}}}\right|\right)$$
Let's take the limit
we find
$$\frac{1}{R} = \tilde{\infty} \left(-5 + \lim_{n \to \infty} \left|{\frac{\cos{\left(\pi n \right)}}{\cos{\left(\pi \left(n + 1\right) \right)}}}\right|\right)$$
$$R = 0 \left(-5 + \lim_{n \to \infty} \left|{\frac{\cos{\left(\pi n \right)}}{\cos{\left(\pi \left(n + 1\right) \right)}}}\right|\right)^{-1}$$
$$\frac{\cos{\left(\pi n \right)}}{5^{n}}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \cos{\left(\pi n \right)}$$
and
$$x_{0} = -5$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-5 + \lim_{n \to \infty} \left|{\frac{\cos{\left(\pi n \right)}}{\cos{\left(\pi \left(n + 1\right) \right)}}}\right|\right)$$
Let's take the limit
we find
$$\frac{1}{R} = \tilde{\infty} \left(-5 + \lim_{n \to \infty} \left|{\frac{\cos{\left(\pi n \right)}}{\cos{\left(\pi \left(n + 1\right) \right)}}}\right|\right)$$
$$R = 0 \left(-5 + \lim_{n \to \infty} \left|{\frac{\cos{\left(\pi n \right)}}{\cos{\left(\pi \left(n + 1\right) \right)}}}\right|\right)^{-1}$$
The rate of convergence of the power series
The answer
[src]
oo ___ \ ` \ -n / 5 *cos(pi*n) /__, n = 1
$$\sum_{n=1}^{\infty} 5^{- n} \cos{\left(\pi n \right)}$$
Sum(5^(-n)*cos(pi*n), (n, 1, oo))
The graph
Examples of finding the sum of a series
- The Sum of the Power Series
x^n/n
(x-1)^n
- Factorial
1/2^(n!)
n^2/n!
x^n/n!
k!/(n!*(n+k)!)
- Flint Hills Series
csc(n)^2/n^3
- Basel problem
1/n^2
1/n^4
1/n^6
- Harmonic series
1/n
- Grandi's series
(-1)^n
- Alternating series
(-1)^(n + 1)/n
(n + 2)*(-1)^(n - 1)
(3*n - 1)/(-5)^n
(-1)^(n - 1)*n/(6*n - 5)
- Newton–Mercator series
(-1)^(n + 1)/n*x^n
- Exam the series for convergence
(3*n - 1)/(-5)^n